Question

Magnesium oxide can be made by heatingmagnesium metal in the presence of oxygen. Thebalanced equation for the reaction is:2 Mg (s) + 02 (g) › 2 MgO (s) When 10.1 gof Mg are allowed to react with 10.5 g of O2, 13.3g of MgO are collected.

Answer 1

To determine the theoretical yield of the reaction we must calculate the moles of reactants that we have. The moles of reactants can be found using the molar mass.

Molar Mass Mg= 24.305g/mol

Molar mass O2=31.998g/mol

Molar mass MgO=40.3044g/mol

Now, the moles of the reactants will be:

`\begin{gathered} molMg=GivengMg*(1molMg)/(MolarMass,gMg) \\ molMg=10.1gMg*(1molMg)/(24.305gMg)=0.41molMg \end{gathered}`
`\begin{gathered} molO_2=GivengO_2*(1molO_2)/(MolarMass,gO_2) \\ molO_2=10.5gMg*(1molO_2)/(31.998gO_2)=0.33molO_2 \end{gathered}`

Now, the calculation of theoretical yield will be with the limiting reactant.

Since by stoichiometry we have that the Mg to O2 ratio is 2/1=2 and the current ratio is 0.41/0.33=1.2. Because the current ratio is less than the theoretical one. In this case, the limiting reactant is Mg.

Now, the moles of MgO that we can obtain using the moles of Mg available will be:

`\begin{gathered} molMgO=givenmolMg*(2molMgO)/(2molMg) \\ molMgO=0.41molMg*(2molMgO)/(2molMg)=0.41molMgO \end{gathered}`

The theoretical yield of MgO (Mass of MgO) will be:

`\begin{gathered} gMgO=givenmolMgO*(MolarMass,gMgO)/(1molMgO) \\ gMgO=0.41molMgO*(40.3044gMgO)/(1molMgO)=16.7gMgO \end{gathered}`

Answer PART B : 16.7 g

Part C

Now, the percent yield will be calculated with the following equation:

Answer PART C: 79.6%